blocks|key|1181356|text|假设S[i]是可以由数字的第一个i组成的数字集。然后给定S[i]，很容易构造S[i%2B1]：从S[i]开始，然后添加s%2Br形式的所有数字，其中s在S[i]中，r是列表中的(i%2B1)-th数字。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1181357|因此，您可以迭代地构建集合s[0]+=+{0},+S[1],...,S[n-r]，并且S[n-r]包含所有可能的和。|1181358|下面是你的例子的结果：|1181359|S[0]+=+{0}
r+=+1:+S[1]+=+{0}+union+{0%2B1}+=+{0,1}
r+=+2:+S[2]+=+{0,1}+union+{0%2B2,1%2B2}+=+{0,1,2,3}
r+=+4:+S[3]+=+{0,1,2,3}+union+{0%2B4,1%2B4,2%2B4,3%2B4}+=+{0,1,2,3,4,5,6,7}|code-block|syntax|javascript|1181360|entityMap^0|2|4|G|1|S|4|12|6|1A|4|1L|3|1Y|1|20|4|26|1|2C|5|0|D|R|17|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1G|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1H|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|P|$]]

Let <code>S[i]</code> be the set of numbers that can be formed from the first <code>i</code> of your numbers. Then given <code>S[i]</code>, it is easy to construct <code>S[i+1]</code>: start with <code>S[i]</code>, and then add all numbers of the form <code>s+r</code>, where <code>s</code> is in <code>S[i]</code> and <code>r</code> is the <code>(i+1)</code>-th number in your list.

Thus you can iteratively build the sets <code>s[0] = {0}, S[1],...,S[n-r]</code>, and <code>S[n-r]</code> contains all possible sums.

Here is how your example plays out:

<pre><code>S[0] = {0}
r = 1: S[1] = {0} union {0+1} = {0,1}
r = 2: S[2] = {0,1} union {0+2,1+2} = {0,1,2,3}
r = 4: S[3] = {0,1,2,3} union {0+4,1+4,2+4,3+4} = {0,1,2,3,4,5,6,7}
</code></pre>

blocks|key|4966786|text|逻辑是为每个连续的整数构建所有可能的和，从1开始。这个问题可以通过跟踪所有可能的和并只检查整数对的和来简化。伪代码(未经测试且有buggy)将如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4966787|const+std::vector<unsigned>+l+=+{1,2,4};
const+unsigned+sum+=+std::accumulate(l.begin(),+l.end());
typedef+std::vector<unsigned>+Sum;+//+one+possibility+for+a+single+value
typedef+std::vector<Sum>+Sums;+//+all+possibilities+for+a+single+value
//+the+element+all[i]+will+provide+all+possible+subsets+where+the+sum+is+equal+to+'i'
std::vector<Sums>+all(sum+%2B+2);+//+we+know+that+sum+%2B+1+is+impossible
//+initialize+with+single+values
for+(auto+i:+l)
{
++++all[i].push_back(Vector(1,+i));
}
unsigned+i+=+1;+//+ignore+0
//+stop+as+soon+as+a+value+doesn't+have+any+subset
while+(!all[i].empty())
{
++++%2B%2Bi;
++++for+(unsigned+j+=+1;+i/2+>+j;+%2B%2Bj)
++++{
++++++++const+unsigned+k+=+i+-+j;
++++++++//+as+i+==+j%2Bk,+create+all+the+relevant+combinations
++++++++for+(const+auto+&sj:+all[j])
++++++++{
++++++++++++for+(const+auto+&sk:+all[k])
++++++++++++{
++++++++++++++++all[i].push_back(sj);
++++++++++++++++all[i].back.insert(all[i].end(),+sk.begin(),+sk.end());
++++++++++++}
++++++++}
++++}
++++if+(0+==+(i+%25+2))
++++{
++++++++//+create+all+the+possible+decompositions+out+of+i/2
++++++++for+(auto+left+=+all[i/2].begin();+all[i/2].end()+!=+left;+%2B%2Bleft)
++++++++{
++++++++++++for+(auto+right+=+left+%2B+1;+all[i/2].end()+!=+right;+%2B%2B+right)
++++++++++++{
++++++++++++++++all[i].push_back(*left);
++++++++++++++++all[i].back.insert(all[i].end(),+right->begin(),+right->end());
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|4966788|需要解决的一个错误是:当相同的数字多次出现时，拒绝求和。|4966789|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The logic is to build all the possible sums for each successive integer, starting from 1.
The problem can be simplified by keeping track of all the possible sums and checking only for sums of pairs of integer. The pseudocode (untested and buggy) would look like:

<pre><code>const std::vector&lt;unsigned&gt; l = {1,2,4};
const unsigned sum = std::accumulate(l.begin(), l.end());
typedef std::vector&lt;unsigned&gt; Sum; // one possibility for a single value
typedef std::vector&lt;Sum&gt; Sums; // all possibilities for a single value
// the element all[i] will provide all possible subsets where the sum is equal to 'i'
std::vector&lt;Sums&gt; all(sum + 2); // we know that sum + 1 is impossible
// initialize with single values
for (auto i: l)
{
 all[i].push_back(Vector(1, i));
}
unsigned i = 1; // ignore 0
// stop as soon as a value doesn't have any subset
while (!all[i].empty())
{
 ++i;
 for (unsigned j = 1; i/2 &gt; j; ++j)
 {
 const unsigned k = i - j;
 // as i == j+k, create all the relevant combinations
 for (const auto &amp;sj: all[j])
 {
 for (const auto &amp;sk: all[k])
 {
 all[i].push_back(sj);
 all[i].back.insert(all[i].end(), sk.begin(), sk.end());
 }
 }
 }
 if (0 == (i % 2))
 {
 // create all the possible decompositions out of i/2
 for (auto left = all[i/2].begin(); all[i/2].end() != left; ++left)
 {
 for (auto right = left + 1; all[i/2].end() != right; ++ right)
 {
 all[i].push_back(*left);
 all[i].back.insert(all[i].end(), right-&gt;begin(), right-&gt;end());
 }
 }
 }
}
</code></pre>

One of the bugs that need to be addressed: rejecting sums where the same number appears more than once.

If I have <code>n-r</code> numbers, from <code>1 to n</code> where <code>r</code> numbers are missing in between, then how can I calculate all possible numbers that can be formed from addition of these numbers (either in groups of 2/3/4/5/6...). 

For example, lets say I have <code>5-2</code> numbers, 
that is, <code>1 2 4</code> and <code>3 5</code> are missing. Now, I can form

<pre><code>1 - {1}
2 - {2}
3 - {1,2}
4 - {4}
5 - {1,4}
6 - {4,2}
7 - {1,2,4}
8 - Cannot be formed
</code></pre>

This is I need to find out, that is the first number from 1 which I cannot form using the combination of the given digits. A simple logic would do fine. Thanks!

All possible numbers formed from addition of given digits

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如果我有来自1 to n的n-r数字，其中r数字在两者之间缺失，那么我如何计算这些数字相加可以形成的所有可能的数字(或者以2/3/4/5/6为一组...)。例如，假设我有5-2号，也就是说，缺少1 2 4和3 5。现在，我可以形成1 - {1}2 - {2}3 - {1,2}4 - {4}5 - {1,4}6 - {4...

问由给定数字相加而形成的所有可能的数字
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